β’ We use Pulse Width Modulation (PWM) to control the voltage magnitude across the load.
β’ For example, in the half bridge circuit we would like to control the output voltage across the load.
β’ This is possible by turning on and off the switches.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 26
β’ In this technique the output voltage is controlled by pulses of voltage.
β’ More is the width of pulse, more is the average voltage in a cycle.
β’ A high frequency carrier (or triangular) wave is compared with a low frequency reference (or modulating) wave. The low frequency reference wave can be DC.
β’ The height of the triangular wave is
assumed 1. The height of the reference wave is assumed to be m.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 27
β’ We follow the logic as:
If mod_wave > tri_wave, S1=on;
else,
S2=on;
β’ What is the average voltage of vAO during the switching cycle Ts?
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 28
β’ From geometry, π΄π΅ π΅π· = π΄πΈ πΈπΆ β’ So, ππ1βπ 2 βπππ = ππ1 2 β’ Thus, π = πππ ππ 2 = duty ratio.
β’ The average voltage of vAO during the switching cycle Ts is:
π£π΄π(ππ£) = ππ· πππππ 2 + 0. ππ 2 βπππ ππ 2 = πππ·
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 29
β’ Note that π£π΄π(ππ£) is the average voltage in a switching cycle TS.
β’ If m varies slowly from cycle to cycle, then the average voltage will also vary from cycle to cycle.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 30
β’ If m varies sinusodially and βslowlyβ over time, then π£π΄π(ππ£) will vary sinusodially with time.
β’ Hence, we can generate a sinusoidal output from the converter.
β’ The triangle frequency should be high enough compared to modulating wave.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 31
β’ What is the expression of output voltage with sine-PWM?
β’ π£π΄π π‘ = π sin ππ‘ ππ·
2 +
ππ· 2
β’ Note that we have used VD/2 as the multiplying factor since sin ππ‘ varies
from +1 to -1.
β’ m usually varies from 0 to 1, sometimes can be more than 1. β’ Additional harmonics are also generated (to be covered later).
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 32
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 33
Linear modulation
β’ The instantaneous value of output voltage (π£π΄π π‘ ) of the half bridge converter always fluctuates between 0 to VD
β’ However, the fundamental output voltage of the converter is linearly
proportional to m. When m varies from 0 to 1 it is called linear modulation region of operation of the half bridge converter.
β’ If m=0, π£π΄π_ππ’ππ_ππ = 0. If m=1, π£π΄π_ππ’ππ_ππ = VD/2 . β’ If m=1, π£π΄π_ππ’ππ_πππ = VD/(2β2)= 0.35VD .
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 34
Over-modulation
β’ If m>1, π£π΄π_ππ’ππ > VD/2 so we can get more output voltage from the converter.
β’ This region of operation is called over-modulation.
β’ However, over-modulation generates lower order harmonics (5th, 7th etc.). Usually overmodulation operation of converter is avoided.
β’ When m becomes very large, square wave operation of the converter is reached. Each switch of the half bridge switches only once in the cycle. β’ We can get maximum output voltage of the half bridge which is given by
π£π΄π_ππ’ππ_ππ_π π = 4
π ππ·
2 = 1.27 times that obtained with sine-PWM.
β’ However, substantial lower order harmonics (5th, 7th etc.) are generated. Usually square wave operation of converter is avoided.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 35
Square wave operation
β’ In a half bridge, π£π΄π_ππ’ππ_ππ = VD/2.
β’ How much voltage can be obtained from a full bridge converter?
β’ π£π΄π΅_ππ’ππ_ππ = π£π΄π_ππ’ππ_ππ β π£π΅π_ππ’ππ_ππ
β’ Leg B can produce a voltage which is 1800 phase shifted from Leg A voltage.
β’ Thus we get double the voltage from a full bridge converter.
β’ π£π΄π΅_ππ’ππ_ππ= VD.
β’ Leg B can be controlled independent of leg A.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 36
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 37
Full bridge circuit waveform
β’ 3 phase circuit is an extension of the half bridge concept.
β’ There are three half bridges working
together. The modulating waveforms for
three half bridges are usually three balanced sine waves having equal amplitude and phase shifted by 1200.
β’ The modulating waveforms for three half bridges need not be balanced.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 38
β’ What is the load voltage? β’ π£π΄π = π£π΄π + π£ππ β’ π£π΅π = π£π΅π + π£ππ β’ π£πΆπ = π£πΆπ + π£ππ β’ Now, (π£π΄π+π£π΅π+π£πΆπ) π = 0 β’ Thus, π£ππ = β (π£π΄π+π£π΅π+π£πΆπ) 3 β’ π£π΄π = 2 3 π£π΄π β 1 3 π£π΅π β 1 3 π£πΆπ β’ π£π΅π = 2 3 π£π΅π β 1 3 π£πΆπ β 1 3 π£π΄π β’ π£πΆπ = 2 3 π£πΆπ β 1 3 π£π΄π β 1 3 π£π΅π
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 39
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 40
3-phase circuit waveform
β’ Let π£π΄π = ππ· 2 (1 + π sin ππ‘ ), π£π΅π = ππ· 2 (1 + π sin(ππ‘ β 2π 3 )), π£πΆπ = ππ· 2 (1 + π sin(ππ‘ β 4π 3 )) β’ Then π£π΄π = 2 3 π£π΄π β 1 3 π£π΅π β 1 3 π£πΆπ = π sin ππ‘ ππ· 2 = π£π΄π
β’ Thus the fundamental voltage across the load is the same as the fundamental pole voltage from the 3 phase converter.
β’ Note that there is always an instantaneous voltage difference (π£ππ) between the load neutral and the negative terminal of the DC bus. So we cannot short them.
β’ Note that VD/2 is the common mode voltage in the output of the converter. There can be other values of common mode voltage also (to be explored later).
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 41
Circuit Peak Pole voltage RMS load voltage RMS line voltage Half Bridge mVD/2 mVD/(2β2)=0.35VD Full Bridge mVD mVD/(β2)=0.7VD 3 phase mVD/2 mVD/(2β2)=0.35VD (per phase) β3mVD/(2β2)=0.6VD
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 42
Load voltage magnitudes
β’ The modulation index m can vary between 0 and 1 with sinusoidal PWM avoiding over modulation operation.
β’ 3-phase RMS load voltage (per phase) can be increased by another 15% by addition of common mode voltage (to be covered later).
β’ From a half bridge converter, we can generate a variable sinusoidal output voltage. β’ π£π΄π π‘ = π sin ππ‘ ππ·
2 +
ππ· 2
β’ What about the harmonics?
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 43
Sinusoidal PWM in half bridge converter
β’ Suppose fc is the carrier frequency and f1 is the modulating frequency.
β’ Harmonics reside around mf , 2mf , 3mf β¦ with sidebands where mf=fc/f1. β’ Sidebands exist at h = jmf Β± k
β’ If j=1, k=2,4,6 etc. β’ If j=2, k=1,3,5 etc.
β’ If j=odd, k=even. β’ If j=even, k=odd.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 44
β’ π£π΄π π‘ = π sin ππ‘ ππ· 2 + ππ· 2 β’ VDC=600V, m = 0.98, so the fundamental voltage is 600*0.5*0.98=294V. β’ mf =21, harmonics reside around mf , 2mf , 3mf β¦ β’ Sidebands exist at (17,19,21,23,25 β¦), (41,43,45,47,39 β¦) and so on. β’ There is a DC of 300V.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 45
β’ What should be the value of mf?
β’ It should be an integer otherwise subharmonics will be produced. So synchronization is needed.
β’ It should be an odd value so that no even harmonics are produced. β’ It should be an odd multiple of 3 to maintain 3 phase symmetry.
β’ The above points are valid only for small values of mf (mf~=21). For large values of mf, we can relax these criteria.
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